Prove that if all squares of all elements are equal identity of group, that it is abelian.
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Expert's answer
2012-08-17T08:03:01-0400
Notice that for any two elements a,b in a group G their commutant [a,b] = a b a^{-1} b^{-1} is generated by squares:
[a,b] = a b a^{-1} b^{-1} = ab ab b^{-1} a^{-1} a^{-1} b^{-1} =
(ab)^2 b^{-1} a^{-1} b b^{-1} a^{-1} b b^{-1} b^{-1} =
(ab)^2 ( b^{-1} a^{-1} b ) ( b^{-1} a^{-1} b ) ( b^{-1} b^{-1}) = (ab)^2 (b^{-1} a^{-1} b)^2 b^{-2}
Hence if all squares of all elements are equal identity of group G, then all commutants in G are identity, i.e. all elements in G commute, that is G is abelian.
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